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Another example of a compact convex set

At the end of my last post on compact convex sets I claimed that every point in one was a countable sum of extreme points. I later realised that I’d made a stronger claim than the theory did, and thus that it was probably false. As such I owe all 5 of you who read my maths posts a counter-example.

One reason it took a little figuring out is that I’m pretty sure the result is true for normed spaces. I haven’t worked out the details, but basically I think you can use a sequence of finite sums approximating it to within \(2^{-n}) to build a countable sum equalling it exactly. I may bother to figure out the details, but I’m about 90% sure they will work.

Let \(V = \mathbb{R}^{2^{\aleph_0}}\) with the product topology. The basic open sets in this are convex, so it’s a locally convex topological vector space.

Let \(A\) be the set of non-negative vectors such that for all \(a, b\), \(x_a + x_b \leq 1\). That is, we’ve “cut off the corners” of the unit cube. This is a closed subset of the cube \([0, 1]^{2^{\aleph_0}}\), which is compact because of Tychonoff’s theorem, therefore is itself compact. It’s an intersection of convex sets, so is also convex.

The extreme points of this are then the origin and all the unit vectors \(e_a\). Therefore any countable sum of extreme points can only be non-zero on countable many coordinates (because convergence is pointwise). But the vector which is uniformly \(\frac{1}{4}\) everywhere is a member of \(A\) which is non-zero at uncountably many coordinates, and thus cannot be a countable sum of the extreme points.